Output of pump required?
from what little i know the total length of the pipe will be at least 600 meters , it could be a 3 or a 4 inch GI pipe. i need to get at least 2000 liters/minute.
i have also been told that the pressure at the end of the hose/pipe should be at 150psi.
so for this what kind of water pump will i require and how many hp will it have to be?
Answer:
HP will depend upon the pump speed you are going to select. Refer QH curve for pumps.
*Head lost due to friction[minimum]
=f*l*v^2/2*g*d
[f = friction factor
L = length of pipe work
D = inner dia of pipe work
v = velocity of fluid
g = acceleration due to gravity]
Add above and vapour pressure at the pumping temperature to the static head.
*Following are the formulae you can use it for calculation as your data are sufficient.
[area=3.14*d^2/4
flow rate=area*velocity
v=sqrt[2*g*h]
[The Colebrook-White formula:
1/sqrt(f) = 1.14 – 2 log10 [e/D + 9.35/(Re x sqrt(f))]
f = friction factor
e = internal roughness of pipe
D = internal diameter of pipe
Re = Reynolds number
Friction factors for turbulent flow calculated by Pipe Flow Expert are based on the Colebrook-White formula.
The friction factor for Laminar flow is calculated from f = 64/Re]
*The fluid head resistance can also be expressed a pressure.
Metric units:
bar = h fluid x p x g / 100000
h = head loss (m)
p = fluid density (kg/m³)
g = acceleration due to gravity (m/s ²)
Imperial units:
psi = h fluid x SG x 2.311
h = head loss (ft)
SG = specific gravity of the fluid
*shaft horse power[HP]
=specific gravity*head available*flow rate]/75
SPECIFIC GRAVITY=1000
Flow rate=.03333 m^3/sec
terminal pressure=150psi
=150/14.22 kg/cm^2
=10.55 kg/cm^2
=105.5 meter of water column
*pump power
=1000*105.5*.033333/75
=46.88 hp[min]
If you are using very high efficient centrifugal pump , then
considering 75%efficiency,
motor hp required will be ,
46.88/0.75
=62.5 hp
*Ultimately its all depends upon your suction condition,ELEVATION and Available npsh.
*Go through following:
*In designing a pumping system, it is essential to provide adequate NPSH available for proper pump operation. Insufficient NPSH available may seriously restrict pump selection, or even force an expensive system redesign. On the other hand, providing excessive NPSH available may needlessly increase system cost.
Suction specific speed may provide help in this situation.
Suction specific speed (S) is defined as:
N*GPM^[1/2]/[NPSHr]^[3/4]
Where
N = Pump speed RPM
GPM = Pump flow at best efficiency point at impeller inlet (for double suction impellers divide total pump flow by two).
NPSHR = Pump NPSH required at best efficiency point.
For a given pump, the suction specific speed is generally a constant - it does not change when the pump speed is changed. Experience has shown that 9000 is a reasonable value of suction specific speed. Pumps with a minimum suction specific speed of 9000 are readily available, and are not normally subject to severe operating restrictions, unless the pump speed pushes the pump into high or very high suction energy.
An example:
Flow 2,000 GPM; head 600 ft. What NPSHA will be required?
Assume: at 600 ft., 3500 RPM operation will be required.
N*GPM^[1/2]/[NPSHr]^[3/4]
9000=3500*2000^0.5/[NPSHr]^[3/...
{9000=suction specific speed}
[NPSHr]^[3/4]=46ft[pump]
NPSHA=[NPSHr]*[NPSH margin ratio]
NPSHA=46*1.5
NPSHA=69ft
A related problem is in selecting a new pump, especially at higher flow, for an existing system. Suction specific speed will highlight applications where NPSHA may restrict pump selection. An example:
Existing system: Flow 2000 GPM; head 600 ft.; NPSHA 30 ft.; Specific Gravity 1.0; Suction Nozzle 6 in. - What is the maximum speed at which a pump can be run without exceeding NPSH available? (NPSHMargin Ratio = 1.5 from above @ S.E. = 173 x 106)
Running a pump at this speed would require a gear and at this speed, the pump might not develop the required head. At a mini-mum, existing NPSH A is constraining pump selection.
Same system as 1. Is a double suction pump practical?
For a double suction pump De = .75 x 6" = 4.5
S.E. = 4.5 x 3550 x 9000 x 1.0
S.E. = 136 x 106 (High S.E.)
For a double suction pump, flow is divided by two.
using the same formula in ex1, we have 2692 RPM]
Using a double suction pump is one way of meeting system NPSH and obtaining a higher head.
no idea
Since you are installing fire hydrant system , you must comply with safety rules by TAC or NFPA etc
Hydrant system pressure for small plants may be limited to 150 psig . Firepumps may be 500 gpm, 150 psi, diesel engine driven . Exact design can only be frozen after getting details of the plant layout, hazard class, equipment to be protected etc.
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